Thanks a lot guys. Really do appreciate it if i could get it before 10:30. It's 7:00 here now. :)

Rule number 6:

Quote:

Do not cheat. Teachers expect questions that form part of an assessment that contributes towards the final grade of a student to be the work of that student and not the work of others. For that reason, MHF policy is to not knowingly help with such questions. If a question presents in such a way as to suggest that it falls in this category, a Moderator will Close the thread (with an explanation as to why the thread was closed). The original poster can always send a pm to the Moderator to discuss the situation but if the Moderator is unconvinced, then the thread will remain closed. When a Moderator is certain that a particular member is attempting to cheat, that member will be banned and the offending thread removed. Where possible, the institute at which the member studies at will also be notified and a copy of the thread provided. This may sound harsh but the fact is that MHF places a high premium on academic honesty and integrity.

July 7th 2011, 03:26 PM

edriann

Re: HELP about Sets please. As soon as possible. (I'm going to need it like 3 1/2 hou

Oops. Sorry. I think I missed that part. So maybe you could just show me how to do it? The professor really didn't teach it. It's not graded anyway, but he said if we don't pass something, we might get scr**ed. Please?

July 7th 2011, 03:38 PM

Also sprach Zarathustra

Re: HELP about Sets please. As soon as possible. (I'm going to need it like 3 1/2 hou

Quote:

Originally Posted by edriann

Oops. Sorry. I think I missed that part. So maybe you could just show me how to do it? The professor really didn't teach it. It's not graded anyway, but he said if we don't pass something, we might get scr**ed. Please?

No, for that, but a little hint, yes.

Use n(AUB)= n(A)+n(B)-n(A∩B).

July 7th 2011, 03:44 PM

edriann

Re: HELP about Sets please. As soon as possible. (I'm going to need it like 3 1/2 hou

Quote:

Originally Posted by Also sprach Zarathustra

No, for that, but a little hint, yes.

Use n(AUB)= n(A)+n(B)-n(A∩B).

I really have no idea how to do that. Hahaha. Anyway, you can please request to close this thread now. Thanks anyway. :)